Calculate the mass of benzoic acid required to prepare 100cm3 each of 0.1m,0.25m, and 0.5m solutions

Answer 1

I assume your m should be Molarity, M.

mass=molartiy*molmass*volumeinLiters
mass=.1*molmassBenzoicAcid*.1m^3
so the mom mass is 122g/mole, so we have
mass=.1 Mole/liter*122g/mole*.1 Liter= 1.22 grams
check my thinking.

Answer 2

M =molar.volume. molar mass÷1000
314×220×1.28÷ 1000
= 0.535